Let’s solve the problem step by step: --- ### Given Series \[ \sum_{n=1}^\infty \frac{1}{n^2} \] --- ### **a) Why is the series convergent? Give a rationale for your answer.** This is a **p-series**: \(\sum_{n=1}^\infty \frac{1}{n^p}\). - A p-series converges if \(p > 1\). - Here, \(p = 2 > 1\). - Therefore, the series converges **by the p-series test**. **Rationale:** For the p-series \(\sum_{n=1}^\infty \frac{1}{n^p}\), if \(p > 1\), the terms decrease fast enough for the series to converge. --- ### **b) Approximate the sum of the series by using the sum of the first 10 terms.** Calculate: \[ S_{10} = \sum_{n=1}^{10} \frac{1}{n^2} \] Let's compute each term: - \(1/1^2 = 1\) - \(1/2^2 = 0.25\) - \(1/3^2 \approx 0.1111\) - \(1/4^2 = 0.0625\) - \(1/5^2 = 0.04\) - \(1/6^2 \approx 0.0278\) - \(1/7^2 \approx 0.0204\) - \(1/8^2 = 0.015625\) - \(1/9^2 \approx 0.0123457\) - \(1/10^2 = 0.01\) Sum: \[ S_{10} \approx 1 + 0.25 + 0.1111 + 0.0625 + 0.04 + 0.0278 + 0.0204 + 0.015625 + 0.0123457 + 0.01 = 1.5497707 \] So, \[ \boxed{S_{10} \approx 1.5498} \] --- ### **c) Estimate the error involved in this estimation.** The error is the sum of the remaining terms: \[ \text{Error} = \sum_{n=11}^\infty \frac{1}{n^2} \] A good estimate uses the **integral test remainder estimate**: \[ \sum_{n=N+1}^\infty \frac{1}{n^2} < \int_{N}^{\infty} \frac{1}{x^2}dx = \left. -\frac{1}{x} \right|_{N}^{\infty} = \frac{1}{N} \] With \(N = 10\): \[ \text{Error} < \frac{1}{10} = 0.1 \] --- ### **d) How many terms are required to ensure that the sum is accurate to within 0.0001?** Set error \(< 0.0001\): \[ \frac{1}{N} < 0.0001 \implies N > 10,000 \] So, **at least 10,001 terms** are required. --- ### **e) Use \( S_n + \int_{n}^{\infty} f(x) dx \leq S \leq S_n + \int_{n+1}^{\infty} f(x) dx \) with \( n = 10 \) to estimate the sum.** Here, \(f(x) = \frac{1}{x^2}\). Calculate the integrals: \[ \int_{10}^{\infty} \frac{1}{x^2} dx = \left. -\frac{1}{x} \right|_{10}^{\infty} = 0 - (-\frac{1}{10}) = 0.1 \] \[ \int_{11}^{\infty} \frac{1}{x^2} dx = \frac{1}{11} \approx 0.09091 \] So, \[ S_{10} + 0.1 \leq S \leq S_{10} + 0.09091 \] \[ 1.5498 + 0.09091 \leq S \leq 1.5498 + 0.1 \] \[ 1.6407 \leq S \leq 1.6498 \] --- ## **Final Answers (Summary Table)** | Part | Answer | |------|--------| | a) | The series converges by the p-series test since \(p = 2 > 1\). | | b) | The sum of the first 10 terms: \(\boxed{1.5498}\) | | c) | Error estimate using the integral: \(< 0.1\) | | d) | At least \(10,001\) terms needed for accuracy within \(0.0001\) | | e) | \(1.6407 \leq S \leq 1.6498\) | **Note:** The exact sum of the series is \(\frac{\pi^2}{6} \approx 1.644934\).